QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#135326 | #6631. Maximum Bitwise OR | BoulevardDust# | RE | 1ms | 79576kb | C++17 | 4.2kb | 2023-08-05 13:47:00 | 2023-08-05 13:47:06 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define out(x) cerr << #x << " = " << x << " "
#define outln(x) cerr << #x << " = " << x << endl
const int N = 100005, K = 29;
int n, q;
int a[N], b[30][N];
int lis0[N], lis1[N], num = 0;
struct seg {
int mx[N << 2], ov[N << 2];
void push_up(int x) {
mx[x] = max(mx[x << 1], mx[x << 1 | 1]);
ov[x] = ov[x << 1] | ov[x << 1 | 1];
}
void build(int x, int l, int r) {
if (l == r) {
mx[x] = ov[x] = a[l];
return;
}
int mid = (l + r) >> 1;
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
push_up(x);
}
int query(int x, int l, int r, int L, int R) {
if (L <= l && r <= R) return mx[x];
int mid = (l + r) >> 1;
if (R <= mid) return query(x << 1, l, mid, L, R);
if (L > mid) return query(x << 1 | 1, mid + 1, r, L, R);
int A = query(x << 1, l, mid, L, R);
int B = query(x << 1 | 1, mid + 1, r, L, R);
return max(A, B);
}
int query_or(int x, int l, int r, int L, int R) {
if (L <= l && r <= R) return ov[x];
int mid = (l + r) >> 1;
if (R <= mid) return query_or(x << 1, l, mid, L, R);
if (L > mid) return query_or(x << 1 | 1, mid + 1, r, L, R);
int A = query_or(x << 1, l, mid, L, R);
int B = query_or(x << 1 | 1, mid + 1, r, L, R);
return A | B;
}
}T0;
struct seg2 {
array<int, 2> mn[N << 2];
void push_up(int x) {
mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
}
void build(int idx, int x, int l, int r) {
if (l == r) {
mn[x] = {b[idx][l], l};
return;
}
int mid = (l + r) >> 1;
build(idx, x << 1, l, mid);
build(idx, x << 1 | 1, mid + 1, r);
push_up(x);
}
void update(int x, int l, int r, int p, int v) {
if (l == r) {
mn[x] = {v, l};
return;
}
int mid = (l + r) >> 1;
if (p <= mid) update(x << 1, l, mid, p, v);
else update(x << 1 | 1, mid + 1, r, p, v);
push_up(x);
}
array<int, 2> query(int x, int l, int r, int L, int R) {
if (L <= l && r <= R) return mn[x];
int mid = (l + r) >> 1;
if (R <= mid) return query(x << 1, l, mid, L, R);
if (L > mid) return query(x << 1 | 1, mid + 1, r, L, R);
array<int, 2> A = query(x << 1, l, mid, L, R);
array<int, 2> B = query(x << 1 | 1, mid + 1, r, L, R);
return min(A, B);
}
}tr[30];
int main() {
int t; scanf("%d", &t); while (t--) {
scanf("%d%d", &n, &q);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
T0.build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= K; ++j) {
b[j][i] = K + 1;
}
vector<int> seq;
for (int j = 0; j <= K; ++j) {
if (a[i] >> j & 1) {
seq.push_back(j);
}
}
b[seq[0]][i] = 0;
for (int j = 1; j < (int)(seq.size()); ++j) {
b[seq[j]][i] = seq[j - 1] + 1;
}
/*for (int j = 0; j <= K; ++j) {
printf("%d %d %d\n", i, j, b[j][i]);
}*/
}
for (int i = 0; i <= K; ++i) {
tr[i].build(i, 1, 1, n);
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int mx = T0.query(1, 1, n, l, r);
int orval = T0.query_or(1, 1, n, l, r);
int H = -1;
for (int i = 0; i <= K; ++i) if (mx >> i & 1) H = i;
if (mx == 0) {
printf("0 0\n");
continue;
}
assert(H >= 0);
int ans = (1 << (H + 1)) - 1;
if (orval == ans) {
printf("%d 0\n", ans);
continue;
}
int pos = -1;
for (int i = 0; i <= K; ++i) {
if (!(orval >> i & 1)) {
pos = i;
break;
}
}
int val = 2;
// out(l); outln(r); outln(orval);
for (int p = H; p >= 0; --p) {
// outln(p);
while (1) {
array<int, 2> tmp = tr[p].query(1, 1, n, l, r);
int far = tmp[0], id = tmp[1];
// out(far); outln(id);
int value = 0;
if (l <= id - 1) value |= T0.query(1, 1, n, l, id - 1);
if (id + 1 <= r) value |= T0.query(1, 1, n, id + 1, r);
// outln(value);
if (far > pos) break;
int w = (1 << (p + 1)) - (1 << far);
if ((value | w) == ans) {
val = 1;
break;
}
lis0[++num] = p;
lis1[num] = id;
tr[p].update(1, 1, n, id, p + 1);
}
for (int i = 1; i <= num; ++i) {
tr[lis0[i]].update(1, 1, n, lis1[i], b[lis0[i]][lis1[i]]);
}
num = 0;
if (val == 1) break;
}
printf("%d %d\n", ans, val);
}
}
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 1ms
memory: 79576kb
input:
1 3 2 10 10 5 1 2 1 3
output:
15 2 15 0
result:
ok 4 number(s): "15 2 15 0"
Test #2:
score: -100
Runtime Error
input:
100000 1 1 924704060 1 1 1 1 149840457 1 1 1 1 515267304 1 1 1 1 635378394 1 1 1 1 416239424 1 1 1 1 960156404 1 1 1 1 431278082 1 1 1 1 629009153 1 1 1 1 140374311 1 1 1 1 245014761 1 1 1 1 445512399 1 1 1 1 43894730 1 1 1 1 129731646 1 1 1 1 711065534 1 1 1 1 322643984 1 1 1 1 482420443 1 1 1 1 20...